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Quadratic Equations: Solving Quadratic Equations Using the Method of Completing the Square

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra</link> by Denny Burzynski and Wade Ellis, Jr. Methods of solving quadratic equations as well as the logic underlying each method are discussed. Factoring, extraction of roots, completing the square, and the quadratic formula are carefully developed. The zero-factor property of real numbers is reintroduced. The chapter also includes graphs of quadratic equations based on the standard parabola, y = x^2, and applied problems from the areas of manufacturing, population, physics, geometry, mathematics (numbers and volumes), and astronomy, which are solved using the five-step method. Objectives of this module: understand the logic underlying the method of completing the square, be able to solve a quadratic equation using the method of completing the square.

Overview

  • The Logic Behind The Method
  • The Method Of Completing The Square

The Logic Behind The Method

Suppose we wish to solve the quadratic equation x 2 −3x−1=0. x 2 −3x−1=0. Since the equation is not of the form x 2 =K, x 2 =K, we cannot use extraction of roots. Next, we try factoring, but after a few trials we see that x 2 −3x−1 x 2 −3x−1 is not factorable. We need another method for solving quadratic equations.
The method we shall study is based on perfect square trinomials and extraction of roots. The method is called solving quadratic equations by completing the square. Consider the equation x 2 +6x+5=0. x 2 +6x+5=0.
This quadratic equation could be solved by factoring, but we’ll use the method of completing the square. We will explain the method in detail after we look at this example. First we’ll rewrite the equation as

 x 2 +6x=−5 x 2 +6x=−5

Then, we’ll add 9 to each side. We get

 x 2 +6x+9=−5+9 x 2 +6x+9=−5+9

The left side factors as a perfect square trinomial.

 ( x+3 ) 2 =4 ( x+3 ) 2 =4

We can solve this by extraction of roots.
 x+3 = + 4 x+3 = ±2 x = ±2−3 x = +2−3 and x=−2−3 x = −1 and −5 x+3 = + 4 x+3 = ±2 x = ±2−3 x = +2−3 and x=−2−3 x = −1 and −5

Notice that when the roots are rational numbers, the equation is factorable.
The big question is, “How did we know to add 9 to each side of the equation?” We can convert any quadratic trinomial appearing in an equation into a perfect square trinomial if we know what number to add to both sides. We can determine that particular number by observing the following situation:

Consider the square of the binomial and the resulting perfect square trinomial

 ( x+p ) 2 = x 2 +2px+ p 2 ( x+p ) 2 = x 2 +2px+ p 2

Notice that the constant term (the number we are looking for) can be obtained from the linear term 2px. 2px. If we take one half the coefficient of x, 2p 2 =p, x, 2p 2 =p, and square it, we get the constant term p 2 . p 2 . This is true for every perfect square trinomial with leading coefficient 1.

In a perfect square trinomial with leading coefficient 1, the constant term is the square of one half the coefficient of the linear term.

Study these examples to see what constant term will make the given binomial into a perfect square trinomial.

Example 1

x 2 +6x. x 2 +6x.   The constant must be the square of one half the coefficient of x. x. Since the coefficient of x x is 6, we have

 6 2 = 3 and 3 2 = 9 6 2 = 3 and 3 2 = 9
The constant is 9.
 x 2 +6x+9= ( x+3 ) 2 x 2 +6x+9= ( x+3 ) 2
This is a perfect square trinomial.

Example 2

a 2 +10a. a 2 +10a.   The constant must be the square of one half the coefficient of a. a. Since the coefficient of a a is 10, we have

 10 2 = 5 and 5 2 = 25 10 2 = 5 and 5 2 = 25
The constant is 25.
 a 2 +10a+25= ( a+5 ) 2 a 2 +10a+25= ( a+5 ) 2

Example 3

y 2 +3y. y 2 +3y.   The constant must be the square of one half the coefficient of y. y. Since the coefficient of y y is 3, we have

 3 2 and ( 3 2 ) 2 = 9 4 3 2 and ( 3 2 ) 2 = 9 4
The constant is 9 4 . 9 4 .
y 2 +3y+ 9 4 = ( y+ 3 2 ) 2 y 2 +3y+ 9 4 = ( y+ 3 2 ) 2

The Method Of Completing The Square

Now, with these observations, we can describe the method of completing the square.

The Method of Completing the Square

  1. Write the equation so that the constant term appears on the right side of equation.
  2. If the leading coefficient is different from 1, divide each term of the equation by that coefficient.
  3. Take one half of the coefficient of the linear term, square it, then add it to both sides of the equation.
  4. The trinomial on the left is now a perfect square trinomial and can be factored as ( ) 2 . ( ) 2 . The first term in the parentheses is the square root of the quadratic term. The last term in the parentheses is one-half the coefficient of the linear term.
  5. Solve this equation by extraction of roots.

Sample Set A

Solve the following equations.

Example 4

x 2 +8x−9 = 0. Add 9 to both sides. x 2 +8x = 9 One half the coefficient of x  is 4, and 4 2  is 16.  Add 16 to both sides. x 2 +8x+16 = 9+16 x 2 +8x+16 = 25 Factor. ( x+4 ) 2 = 25 Take square roots. x+4 = ±5 x = ±5−4 +5−4=1,−5−4=−9 x = 1, −9 x 2 +8x−9 = 0. Add 9 to both sides. x 2 +8x = 9 One half the coefficient of x  is 4, and 4 2  is 16.  Add 16 to both sides. x 2 +8x+16 = 9+16 x 2 +8x+16 = 25 Factor. ( x+4 ) 2 = 25 Take square roots. x+4 = ±5 x = ±5−4 +5−4=1,−5−4=−9 x = 1, −9

Example 5

x 2 −3x−1 = 0. Add 1 to both sides. x 2 −3x = 1 One half the coefficient of x is  −3 2 .  Square it:  ( −3 2 ) 2 = 9 4 . Add  9 4  to each side. x 2 −3x+ 9 4 = 1+ 9 4 x 2 −3x+ 9 4 = 13 4 Factor. Notice that since the sign of the middle  term of the trinomial is "−", its factored form has a "−" sign. ( x− 3 2 ) 2 = 13 4 Now take square roots. x− 3 2 = ± 13 4 x− 3 2 = ± 13 2 x = ± 13 2 + 3 2 x = ± 13 +3 2 x = 3± 13 2 x 2 −3x−1 = 0. Add 1 to both sides. x 2 −3x = 1 One half the coefficient of x is  −3 2 .  Square it:  ( −3 2 ) 2 = 9 4 . Add  9 4  to each side. x 2 −3x+ 9 4 = 1+ 9 4 x 2 −3x+ 9 4 = 13 4 Factor. Notice that since the sign of the middle  term of the trinomial is "−", its factored form has a "−" sign. ( x− 3 2 ) 2 = 13 4 Now take square roots. x− 3 2 = ± 13 4 x− 3 2 = ± 13 2 x = ± 13 2 + 3 2 x = ± 13 +3 2 x = 3± 13 2

Example 6

3 a 2 −36a−39 = 0. Add 39 to both sides. 3 a 2 −36a = 39 The leading coefficient is 3 and we  need it to be 1. Divide each term by 3. a 2 −12a = 13 One half the coefficient of a is −6.  Square it:  ( −6 ) 2 =36. Add 36 to each side. a 2 −12a+36 = 13+36 a 2 −12a+36 = 49 ( a−6 ) 2 = 49 Factor. a−6 = ±7 a = ±7+6 +7+6=13,−7+6=−1 a=13, −1 3 a 2 −36a−39 = 0. Add 39 to both sides. 3 a 2 −36a = 39 The leading coefficient is 3 and we  need it to be 1. Divide each term by 3. a 2 −12a = 13 One half the coefficient of a is −6.  Square it:  ( −6 ) 2 =36. Add 36 to each side. a 2 −12a+36 = 13+36 a 2 −12a+36 = 49 ( a−6 ) 2 = 49 Factor. a−6 = ±7 a = ±7+6 +7+6=13,−7+6=−1 a=13, −1

Example 7

2 x 2 +x+4 = 0 2 x 2 +x = −4 x 2 + 1 2 x = −2 x 2 + 1 2 x+ ( 1 4 ) 2 = −2+ ( 1 4 ) 2 ( x+ 1 4 ) 2 =−2+ 1 16 = −32 16 + 1 16 = −31 16 2 x 2 +x+4 = 0 2 x 2 +x = −4 x 2 + 1 2 x = −2 x 2 + 1 2 x+ ( 1 4 ) 2 = −2+ ( 1 4 ) 2 ( x+ 1 4 ) 2 =−2+ 1 16 = −32 16 + 1 16 = −31 16
Since we know that the square of any number is positive, this equation has no real number solution.

Example 8

Use a calculator. Calculator problem.  Solve 7 a 2 −5a−1=0. 7 a 2 −5a−1=0. Round each solution to the nearest tenth.
 7 a 2 −5a−1 = 0 7 a 2 −5a = 1 a 2 − 5 7 a = 1 7 a 2 − 5 7 a+ ( 5 14 ) 2 = 1 7 + ( 5 14 ) 2 ( a− 5 14 ) 2 = 1 7 + 25 196 = 28 196 + 25 196 = 53 196 a− 5 14 = ± 53 196 = ± 53 14 a= 5 14 ± 53 14 = 5± 53 14 7 a 2 −5a−1 = 0 7 a 2 −5a = 1 a 2 − 5 7 a = 1 7 a 2 − 5 7 a+ ( 5 14 ) 2 = 1 7 + ( 5 14 ) 2 ( a− 5 14 ) 2 = 1 7 + 25 196 = 28 196 + 25 196 = 53 196 a− 5 14 = ± 53 196 = ± 53 14 a= 5 14 ± 53 14 = 5± 53 14

  1. We will first compute the value of the square root.

     Type 53 Press x Display reads: 7.2801099 Type 53 Press x Display reads: 7.2801099
    Press the key that places this value into memory.
  2. For a= 5+ 53 14 , a= 5+ 53 14 ,

    Type 5 Press + Press the key that recalls the value in memory. Press = Press ÷ Type 14 Press = Display reads: .87715071 Type 5 Press + Press the key that recalls the value in memory. Press = Press ÷ Type 14 Press = Display reads: .87715071
    Rounding to tenths, we get a≈0.9. a≈0.9.
  3. For a= 5− 53 14 a= 5− 53 14

    Type 5 Press − Press the key that recalls the value in memory. Press = Press ÷ Type 14 Display reads: −.16286499 Type 5 Press − Press the key that recalls the value in memory. Press = Press ÷ Type 14 Display reads: −.16286499
    Rounding to tenths, we get a≈−0.2. a≈−0.2. Thus, a≈0.9 a≈0.9 and −0.2 −0.2 to the nearest tenth.

Practice Set A

Solve each of the following quadratic equations using the method of completing the square.

Exercise 1

x 2 −2x−48=0 x 2 −2x−48=0

Solution

x=−6, 8 x=−6, 8

Exercise 2

x 2 +3x−5=0 x 2 +3x−5=0

Solution

x= −3± 29 2 x= −3± 29 2

Exercise 3

4 m 2 +5m=−1 4 m 2 +5m=−1

Solution

m= −1 4 , −1 m= −1 4 , −1

Exercise 4

5 y 2 −2y−4=0 5 y 2 −2y−4=0

Solution

y= 1± 21 5 y= 1± 21 5

Exercise 5

Use a calculator. Calculator problem.  Solve 3 x 2 −x−1=0. 3 x 2 −x−1=0. Round each solution to the nearest tenth.

Solution

x=0.8, −0.4 x=0.8, −0.4

Exercises

For the following problems, solve the equations by completing the square.

Exercise 6

x 2 +2x−8=0 x 2 +2x−8=0

Solution

x=−4, 2 x=−4, 2

Exercise 7

y 2 −5y−6=0 y 2 −5y−6=0

Exercise 8

a 2 +7a+12=0 a 2 +7a+12=0

Solution

a=−3, −4 a=−3, −4

Exercise 9

x 2 −10x+16=0 x 2 −10x+16=0

Exercise 10

y 2 −2y−24=0 y 2 −2y−24=0

Solution

y=−4, 6 y=−4, 6

Exercise 11

a 2 +2a−35=0 a 2 +2a−35=0

Exercise 12

x 2 +2x+5=0 x 2 +2x+5=0

Solution

No real number solution.

Exercise 13

x 2 −6x+1=0 x 2 −6x+1=0

Exercise 14

x 2 +4x+4=0 x 2 +4x+4=0

Solution

x=−2 x=−2

Exercise 15

a 2 +4a+7=0 a 2 +4a+7=0

Exercise 16

b 2 +5b−3=0 b 2 +5b−3=0

Solution

b= −5± 37 2 b= −5± 37 2

Exercise 17

b 2 −6b=72 b 2 −6b=72

Exercise 18

a 2 +10a−9=0 a 2 +10a−9=0

Solution

a=−5± 34 a=−5± 34

Exercise 19

a 2 −2a−3=0 a 2 −2a−3=0

Exercise 20

x 2 −10x=0 x 2 −10x=0

Solution

x=10, 0 x=10, 0

Exercise 21

y 2 −8y=0 y 2 −8y=0

Exercise 22

a 2 −6a=0 a 2 −6a=0

Solution

a=6, 0 a=6, 0

Exercise 23

b 2 +6b=0 b 2 +6b=0

Exercise 24

− x 2 −14x=13 − x 2 −14x=13

Solution

x=−13, −1 x=−13, −1

Exercise 25

− x 2 +8x=−84 − x 2 +8x=−84

Exercise 26

2 a 2 +2a−1=0 2 a 2 +2a−1=0

Solution

a= −1± 3 2 a= −1± 3 2

Exercise 27

4 b 2 −8b=16 4 b 2 −8b=16

Exercise 28

9 x 2 +12x−5=0 9 x 2 +12x−5=0

Solution

x= 1 3 ,− 5 3 x= 1 3 ,− 5 3

Exercise 29

16 y 2 −8y−3=0 16 y 2 −8y−3=0

Exercise 30

2 x 2 +5x−4=0 2 x 2 +5x−4=0

Solution

x= −5± 57 4 x= −5± 57 4

Exercise 31

3 a 2 +2a−24=0 3 a 2 +2a−24=0

Exercise 32

x 2 +2x+8=0 x 2 +2x+8=0

Solution

No real number solution.

Exercise 33

y 2 −3y+10=0 y 2 −3y+10=0

Exercise 34

7 a 2 +3a−1=0 7 a 2 +3a−1=0

Solution

a= −3± 37 14 a= −3± 37 14

Use a calculator. Calculator Problems

For the following problems, round each solution to the nearest hundredth.

Exercise 35

5 m 2 −2m−6=0 5 m 2 −2m−6=0

Exercise 36

3 y 2 +5y=7 3 y 2 +5y=7

Solution

y=0.91, −2.57 y=0.91, −2.57

Exercise 37

1.8 x 2 +2.3x−4.1=0 1.8 x 2 +2.3x−4.1=0

Exercise 38

0.04 a 2 −0.03a+0.02=0 0.04 a 2 −0.03a+0.02=0

Solution

No real number solution.

Exercises For Review

Exercise 39

((Reference)) Factor 12ax−6bx+20ay−10by 12ax−6bx+20ay−10by by grouping.

Exercise 40

((Reference)) Graph the compound inequality −6≤−2x+2<−4. −6≤−2x+2<−4.

A horizontal line with arrows on both ends.

Solution

A number line with arrows on each end, and labeled from negative two to five in increments of one. There is a closed circle at four and an open circle at three. These circles are connected by a black line.

Exercise 41

((Reference)) Find the equation of the line that passes through the points ( 1,−2 ) ( 1,−2 ) and ( 0,4 ). ( 0,4 ).

Exercise 42

((Reference)) Find the product: x 2 −4x−12 x 2 −2x−8 ⋅ x 2 −3x−4 x 2 −3x−18 . x 2 −4x−12 x 2 −2x−8 ⋅ x 2 −3x−4 x 2 −3x−18 .

Solution

x+1 x+3 x+1 x+3

Exercise 43

((Reference)) Use the method of extraction of roots to solve ( x−2 ) 2 =25. ( x−2 ) 2 =25.

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